Problem from mba.com practice test 1

I dare to say that the answer is C.

Rewrite the equation in form of quadratic
y=x^2+(a+b)x+ab

0=x^2+(a+b)x+ab - real solutions of this equation are x-coordinates of points where the graph intersects the x-axis.

(1)
a+b=-1 => x^2-x+ab=0
a+b=-1 - does not give enough data to be sure about ab
INSUFFICIENT

(2) (0,6) - this is one root
0=x^2+(a+b)x+ab; (x-6)(x+z)=0; x2+zx-6x-6z=0; x2+(z-6)x-6z=0
there is no enough information to solve this equation further
INSUFFICIENT

(3) COMBINE
from (2) x2+(z-6)x-6z=0
from (1) 0=x^2+(a+b)x+ab; a+b=-1
=>(a+b)=z-6=-1 => z=5 => -6z=-30
hence the equation sought is x2-x-30=0 or (x-6)(x+5)=0
thus, roots of (0,6) and (0,-5) are coordinates of points where the graph intersects the x-axis.
SUFFICIENT

I'll be appreciate someone who point out a mistake in my solution or show a shorter/simpler way.

Best!